What is the proof for the lower and upper Riemann integral?

[twoflower]

Hi all, I'm learning for my analysis exam and I encountered a theorem the proof of which I don't fully understand.

Here it is:

Form of lower and upper Riemann integral

$$\mbox{Let } f \mbox{ be bounded on } [a,b] \mbox{ and let } \left\{D_n\right\}_{n=1}^{\infty} \mbox{ be infinite sequence of dividings of interval } [a,b]. \mbox{ Let } \lim_{n \rightarrow \infty} \Vert D_n \Vert = 0. \mbox{ Then \\ }$$
$$(R) \int_{a}^b f(x)\ dx = \sup_{n} (f, D_n) \mbox{ (lower Riemann integral)}$$
$$(R) \int_{a}^b f(x)\ dx = \inf_{n} S(f, D_n) \mbox{ (upper Riemann integral)}$$

$$\mbox{ Where } s(f, D) \mbox{ and } S(f, D) \mbox{ are lower and upper sums.}$$

Proof:

Let's choose dividing D and $\epsilon > 0$. It's sufficient to prove, that

$$\exists \ n_0 \in \mathbb{N}: s(f, D_{n_0}) > s(f, D) \ - \ \epsilon$$

Because then

$$\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}: \sup_{D'} s(f, D') \ge \sup_{n} s(f, D_n) \ge s(f, D) \ - \ \epsilon$$

We have fixed D, $\epsilon$ and let

$$K = \sup_{x \in [a,b]} |f(x)|$$

We choose $n_0$ such, that

$$\Vert D_{n_0} \Vert < \frac{\epsilon}{K.\sharp D'}$$, where $\sharp D'$ = number of intervals in D.

Let

$$P = D_{n_0} \cup D.$$

$$s(f, D) \le s(f, P) = \sum_{I \in P} (\inf_{I} f).|I|$$ (I doesn't contain points of D)

$$+ \sum_{I \in P} (\inf_{I} f).|I|$$ (I contains at least 1 point from D)

$$\le s(f, D_{n_0}) + K.\Vert D_{n_0} \Vert .\sharp D$$

$$< s(f, D_{n_0}) + \epsilon \ \ \Box$$

Could you please clarify the main idea and individual steps to me?

 

[matt grime]

.

1. f is assumed Riemann integrables

what part do you not understand?

Proof:

Let's choose dividing D and $\epsilon > 0$. It's sufficient to prove, that

$$\exists \ n_0 \in \mathbb{N}: s(f, D_{n_0}) > s(f, D) \ - \ \epsilon$$

ie show the sup of the lower sums of D_n is a sup for the set of all partitions.

Because then

$$\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}: \sup_{D'} s(f, D') \ge \sup_{n} s(f, D_n) \ge s(f, D) \ - \ \epsilon$$

and thus the sup over all D must equal the sup over the D_n

what follows is a straight forward estimate of the lower sums

We have fixed D, $\epsilon$ and let

$$K = \sup_{x \in [a,b]} |f(x)|$$

We choose $n_0$ such, that

$$\Vert D_{n_0} \Vert < \frac{\epsilon}{K.\sharp D'}$$, where $\sharp D'$ = number of intervals in D.

Let

$$P = D_{n_0} \cup D.$$

$$s(f, D) \le s(f, P) = \sum_{I \in P} (\inf_{I} f).|I|$$ (I doesn't contain points of D)

$$+ \sum_{I \in P} (\inf_{I} f).|I|$$ (I contains at least 1 point from D)

$$\le s(f, D_{n_0}) + K.\Vert D_{n_0} \Vert .\sharp D$$

$$< s(f, D_{n_0}) + \epsilon \ \ \Box$$

Could you please clarify the main idea and individual steps to me?

 

[steven187]

Hi all, I'm learning for my analysis exam and I encountered a theorem the proof of which I don't fully understand.

Here it is:

Form of lower and upper Riemann integral

$$\mbox{Let } f \mbox{ be bounded on } [a,b] \mbox{ and let } \left\{D_n\right\}_{n=1}^{\infty} \mbox{ be infinite sequence of dividings of interval } [a,b]. \mbox{ Let } \lim_{n \rightarrow \infty} \Vert D_n \Vert = 0. \mbox{ Then \\ }$$
$$(R) \int_{a}^b f(x)\ dx = \sup_{n}s(f, D_n) \mbox{ (lower Riemann integral)}$$
$$(R) \int_{a}^b f(x)\ dx = \inf_{n} S(f, D_n) \mbox{ (upper Riemann integral)}$$

$$\mbox{ Where } s(f, D) \mbox{ and } S(f, D) \mbox{ are lower and upper sums.}$$

Proof:

Let's choose dividing D and $\epsilon > 0$. It's sufficient to prove, that

$$\exists \ n_0 \in \mathbb{N}: s(f, D_{n_0}) > s(f, D) \ - \ \epsilon$$
$$\exists \ n_0 \in \mathbb{N}:s(f, D) \ - s(f, D_{n_0}) < \epsilon$$ by knowing that the more points occupied in the partition will increase the lower sum so obviously infinitly partitioned lower sum minus the finite partition of the lower sum is less than epsilon, the rest is just displaying an estimate of the lower sums by the looks of it, don't know what else to say
?

......

 

[twoflower]

I understand the first inequality, but I don't understand this:

$$\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}: \sup_{D'} s(f, D') \ge \sup_{n} s(f, D_n) \ge s(f, D) \ - \ \epsilon$$

Why is there D', for what?

 

[matt grime]

You'retaking the sup over all partitions D', ie "the lower riemann integral"

 

[twoflower]

You'retaking the sup over all partitions D', ie "the lower riemann integral"

Well, however I don't see why couldn't I just write

$$\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}: \sup_{n} s(f, D_n) \ge s(f, D) \ - \ \epsilon$$

 

[matt grime]

You can't because that is what you need to prove, since you do not know the sup over the countable family D_n is an upper bound for any D - this is what the proof establishes. There is no need for the there exists n_0 part as well

 

[steven187]

$$\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}:s(f, D) \ - \sup_{D'} s(f, D') \le s(f, D) \ -\sup_{n} s(f, D_n) \le \epsilon$$

for the first inequality, the more intervals (D') occupied in the partition will increase the lower sum so obviously infinitly partitioned lower sum minus the finite partition of the lower sum is less than epsilon,
for the second inequality the more points (n) occupied in the partition will increase the lower sum so obviously infinitly partitioned lower sum minus the finite partition of the lower sum is less than epsilon, well this is basically what i think its sayin by that expression don't know what else to say sorry

 

[matt grime]

typionfg one handed please bear with me


Can I just explain what ther prrof is trying to do rather than hust deconstructing it as it stands.

We want to find the Sup of the lower sums takrn ove all partitions as the length of the largest subinterval tends to zero. this is of course impossible to do in almost any real situation. instead we prove this result that states if we can find a countable family D_n where the sup exists here and |d_n| tends tpo zero then that is sufficient.

obviusly the sup over all partitions is greater than the sup over the family D_n.

We must show that the reverse is true. so take an arbitray partition, D and we show that we can makr the s(f,D_n) greater than s(f,D)-e for all e for all an suff large. take sups of both sides as the sze of the subinterval tendss to zero and we get

sup(f,D_n)=>sup(f,D)

as required

 

[twoflower]

Thank you matt, I think I finally have it. One more:

why (in the second inequality) is there "..exists n_0..." when we don't use it at all? (we're taking sup over all n).

 

[matt grime]

yeah, but the n_0 doesn't appear at all in the statement that follows it does it?

or if you like, remove the "there exitst n_0", does it change the statement at all? no, all variables are quantified properly - it is an unnecessary quantifier

 

[twoflower]

yeah, but the n_0 doesn't appear at all in the statement that follows it does it?

In the proof later it does appear and it is because the first inequality (the one we're acually proving) is stated in a such way that it contains n_0. But the second inequality (which is consequence of the first one) doesn't contain n_0 and it seems useless to me there...

 

[twoflower]

yeah, but the n_0 doesn't appear at all in the statement that follows it does it?

or if you like, remove the "there exitst n_0", does it change the statement at all? no, all variables are quantified properly - it is an unnecessary quantifier

Ok, thank you for the explanation.